The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. Ho
1 answer:
Answer:
Width of sound beam is 7.557 m
Explanation:
First we will calculate the wave length from given data:
λ=v/f
Were:
v is the speed
f is the frequency
We considered the opening long and narrow, Using single slit diffraction formula:
mλ=dsinΘ
where:
d is the crack width
m is the order
Θ is angle
Considering m=1, The angle between first minimum from center of beam is:
The width of beam is:
tanΘ=y/L
Width=7.557 m
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11)
12)
13) 50.0 m/s
14) 41.6 m/s
Explanation:
11)
The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by
where
m is the mass of the object
g is the acceleration due to gravity
h is the height relative to the ground
Here in this problem, when the train is at the top, we have:
m = 8325 kg (mass of the train + riders)
(acceleration due to gravity)
h = 127 m (height of the train at the top)
Substituting,
12)
According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:
where
is the kinetic energy at the top
is the potential energy at the top
is the kinetic energy at the bottom
is the potential energy at the bottom
The kinetic energy is the energy due to motion; since the train is at rest at the top, we have
Also, at the bottom the height is zero, so the potential energy is zero
Therefore, we find:
13)
The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by
where
m is the mass of the object
v is the speed of the object
From question 12), we know that the kinetic energy of the train at the bottom is
We also know that the mass is
m = 8325 kg
Therefore, we can calculate the speed of the train at the bottom:
14)
At the top of the second hill, the total mechanical energy of the train is still conserved.
Therefore, we can write again:
where
is the kinetic energy at the top of the 1st hill
is the potential energy at the top of the 1st hill
is the kinetic energy at the top of the 2nd hill
is the potential energy at the top of the 2nd hill
From the previous questions, we know that
and
The height of the second hill is
h = 39 m
So we can also find the potential energy at the second hill:
So, the kinetic energy at the second hill is
And so, the speed is